Find the angle between force ` vec F =(3 hat I + 4 hat j -5 hat k)` and displacement ` vec d =(5 hat I + 4 hat j + 3 hat k)` unit. Also find the projection of ` vec F and vec d`.

Explanation

Here, `vec F = (3 hat I + 4 hat j - 5 khat k) ,`
`verc d = (5 hat I + 4 hat j + 3 hat k) `
` vec F . Vec D =(3 hat I + 4 hat j -5 hat k) . (5 hat I + 4 hat j + 3 hat k)`
`= 3 (5) + 4 (4) -5 93) = 16 units
`F= sqrt ( F_(_(x)^(2) + F_(y)^(2) + F_(z)^(2)) `
` = sqrt (3^(2) + 4^(2) + (-5)^(2) ) =sqrt 50`
`d= sqrt (d_(x)^(2) + d_(y) ^(2) + d_(z)^(2) ) = sqrt (5^(2) + 4^(2) + 3^(3) ) =sqrt 50)`
Now , cos theta = (vec F. vec d) /(F d) = (16)/(sqrt 50 dqrt 50 )) =(16) /(50) =0.32`
`thets = cos ^(-1) (0. 32) = 71 .3^^(@)`
Unit vector along `vec d ` is
`vec d = (5 hat i + 4 hat j _ 3 hat k)/(sqrt ( 5^(2) +4 %^(2) + 3^(2))) =(5 hat i + 4 hat j + 3 hatk)/(sqrt 50 `
` vec F . vec d = (3 hat i _ 4 hat j - 5 hat k) . (( 5 hat i + 4 hat j + 3 hat k)/(sqrt 50 )`
`=(3 (5) + (4) -5 (3)) /(sqrt 50 ) = (16) /(sqrt)` :. Projection of `vec F on vec d` = component vector of `vec F ` along vedc d` In is`
` =( vec F .vec d) vec d = (160 /(sqrt 50 ) ((5 hat i + 4 hat j + 3 hat k)) /(sqrt50)`
` =0. 32 (5 hat i +4 hat j+ 3 hat k)`.